Homework 4 Due Thursday Oct 7 CLRS 12-4 (number of binary trees) - - PDF document

Homework 4 Due Thursday Oct 7

• CLRS 12-4 (number of binary trees)
• CLRS 13.3-6 (rb insert implementation)

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Chapter 13: Red-Black Trees A red-black tree is a node-colored BST. Each node is colored either black or red. The following special rules apply:

• 1. The root is always black.
• 2. A nil is considered to be black. This

means that every non-NIL node has two children.

• 3. Black Children Rule: The children of

each red node are black.

• 4. Black Height Rule: For each node v,

there exists an integer bh(v) such that each downward path from v to a nil has exactly bh(v) black real (i.e. non-nil)

• nodes. Call this quantity the black

height of v. We define the black height of an RB tree to be the black height of its root.

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An Example

NIL NIL

: black nodes

NIL NIL NIL NIL NIL NIL NIL

: red nodes 2 8 3 11 9 14 15 10

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RB Trees Are Balanced Lemma A Let T be an RB tree having some n ≥ 1 nodes. Then the height of T is at most 2 lg(n + 1) − 1. Proof Let h be the height of T. Let v0, v1, . . . , vh+1 be an arbitrary length h downward path from the root to a nil, where v0 is the root, vh is the leaf, and vh+1 is the

• nil. v0 is black and vh+1 is black. The

number of red nodes among v1, . . . , vh is maximized when for all odd i vi is red. So, the number of red nodes is at most h/2. This means that the number of black ones among v1, . . . , vh is at least h − h/2. Thus, bh(T) ≥ h/2

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Proof (cont’d) If a node has a real black child, then it has another child. This means that the tree contains a complete binary tree of height bh(T) − 1, consisting solely of real black

• nodes. The number of nodes in the complete

binary tree is 2bh(T) − 1 black nodes in T. This number is at most n. So, we have lg(n + 1) ≥ bh(T) ≥ h/2. By solving this we have h ≤ 2 lg(n + 1).

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Operations on RB Trees We will study two operations, insertion and

• deletion. The two operations make use of two
• perations, Left-Rotate and Right-Rotate.

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Left-Rotate and Right-Rotate

right-rotate at y left-rotate at x α β γ x β γ x α y y

Rotations do not break the BST-property.

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• 1. Insertion

Suppose that we want to insert a node x into an RB tree T. To do this, we insert x as a red node using the insertion algorithm for BST’s and then resolve violation of the coloring rules. The exception is when the tree T is empty. Then we color x black. Will this operation violate any rule?

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Will this operation violate any rule? Because the node x is red, the Black Children Rule may be violated. The violation happens when the parent of x is red. The other rules are not violated.

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Enforcing the Black Children Rule After Insertion Let p be the parent of x. Assume p is red;

• therwise there is no violation. Since p is red,

it cannot be the root. So, let g be the grand parent of x. Let u be the sibling of p. Let s be the sibling of x, which can be nil. Since there was no violation before insertion of x, g is black and s is black. However, u can be either red or black. We assume that p is the left child of g. The treatment in the case when p is the right child is similar. We consider two cases, u is black and u is red.

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(Case 1) color[u] = black First, if x is not the left child of p, then left-rotate at p and swap the role of x and that of p. This preserves bh at g’s position. Next, right-rotate at g, then swap the color

• f p and that of g. Also, at the end the node

at g’s position is black. So, there is no violation any more.

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u u e g d s e p s x d g

new−x new−p new−s left−rotate at p new−d new−e

g u x u s d e s d g e

swap colors

• f p & g

d p p x u g x p s e

right−rotate at g violation resolved 12

(Case 2) color[u] = red In this case, we color both p and u black and color g red. This eliminates the violation of the Black Children Rule between p and x, but may introduce violation, which is between g and its parent. So, we may be back to square

• ne, but the location of the violation, if

introduced, is two levels closer to the root. Thus, the bad situation does not repeat more than the height of the tree.

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u p g s x u p g s x violation may occur

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What will happen at the end? Either Case 1 holds or the resolution for Case 2 does not introduce violation. What other condition might be violated?

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It is when the resolution for Case 2 eliminates violation of the Black Children Rule but turns the root red.

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Example 1:

4 5 7 8 2 3 11 14 15 13 p x u g case 2 applies 4 5 7 8 2 3 11 14 15 13 p x s u g case 1 applies 11 14 15 13 8 s x p u g 3 7 5 2 4 13 15 8 11 7 3 5 2 4 14 swap colors 13 15 8 11 7 3 5 2 4 14

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Example 2:

10 20 1 15 12 14 17

swap colors left−rotate

10 20 1 14 17 15 12

right−rotate

10 1 15 14 17 20 12

left−rotate

10 15 20 1 14 17 12

swap colors

20 1 14 17 10 15 12

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• 2. Deletion of a node z

We first apply BST-Delete. In the case when a node is copied to z (the successor of z comes to z’s position), color the new one by the color of z. The deletion routine gives back the pointer, x, to a node where the actual elimination took places. There are two possibilities: (1) There was a leaf at x’s position and x is a nil (2) The node who was at x’s position had a unique child and now this unique child is at x’s position. In the latter case, the unique child is red. Why?

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In the latter case, the unique child is red. Why? Assume otherwise. Then the black height of the

• ther subtree, which is a nil,

is one, while the black height

• f the unique subtree is at

least two. That means that the Black Height Rule is already violated.

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So, the latter case will not create violation. Thus, we will consider only the case when there was a leaf at x is position. Furthermore, if the leaf that has been eliminated is red, then the elimination does not introduce violation. So, we assume that the leaf is black.

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Resolving the Black Height Rule Violation Let w be a node. We say that Few(w) holds if the following conditions are satisfied:

• The Black Height Rule is violated in the

red-black tree.

• Tw, the subtree rooted at w, is a red-black

tree without coloring rule violation, except the rule about the color of the root.

• If bh(w) were bh(w) + 1 then the

Black-Height Rule would be satisfied for all the nodes outside Tw

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Some Properties of Few(·)

• The condition Few initially holds at the

nil who’s replacing x.

• If Few(w) holds and w is red, then

coloring w black resolves violation. So, we will consider the case when w is black.

• If Few(w) holds then w cannot be the

root.

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Assume that x is the left child of its parent p. The case in which x is the right child can be solved similarly. Let s be x’s sibling. Let A and B be the left child and the right child of s, respectively. These two nodes exist since we’ve eliminated a real black node. Here we can preprocess the trees so that the following two conditions are satisfied:

• 1. the sibling of x is black and
• 2. if the left child of the sibling of x is

red, then the right child of the sibling

• f x is red.

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x B A

• 1

s p

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Establishing Property #1 We need to make sure that the sibling of x is black. Suppose that s is red. Then its parent and its children are all black. That is, p, A, and B are black. Suppose we left-rotate at p and then swap the colors of p and s. Then,

• the sibling of x is now A and is black;
• the bh-value is unchanged for x, A, and

B, so Few(x) still holds;

• the depth of x, i.e. the location of Few, is

increased by 1; and

• the parent of x is red.

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−1 B A p x swap colors x A p B −1 s left−rotate s

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Establishing Property #2 We need to ensure that if the left child of the sibling of x is red, then the right child of the sibling of x is red. Suppose that A is red and B is black. Let C and D be the children of A. Then these two are black. Suppose we right-rotate at s and swap the colors of A and s. Then,

• A becomes the sibling of x;
• C and s become the left child and the

right child of A, respectively;

• the bh-value is preserved for x, C, D, B,

and the sibling of x, so Few(x) still holds;

• the left child of x’s sibling is now black.

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C D B x A s −1 p B A s C D x −1 p right− rotate A −1 C D s B x −1 p swap colors

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Resolution After Enforcing the Conditions In the case when A and B are both black, we do the following. If p is black, then we color s red. Then, the downward paths going through s lose one black node. Thus, the location of the Few condition moves one level up to p. If p is red, we color p black and color s red. Then the Few condition disappears.

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s x A B p s x A B p s p x A B p s x A B

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The Remaining Cases p A B status B B B done R B B done R R B prohibited B R B prohibited R R R to be done R B R to be done B R R to be done B B R to be done We left-rotate at p. Furthermore, in the case

• f (R,R,R) we color s red and color p and B

black and in the case of (B,R,R) and (B,B,R) we color B black.

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B B A A A p p s s x x s x B p B B p p s x x A s A B B A A A s x p x p s x B p s B B s x p x A p s x A B p s A

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